Vimeo API集成

Question

我想創建一個網站，使用viemo api來搜索視頻，然後以XML格式輸出它們。我開局良好，但現在我陷入困境。用戶輸入被髮布，但沒有結果顯示Firebug控制檯上了......

這裏是我的VIMEO API集成.PHP：

<? 
include('connect.php'); 

$video_id= $_POST['text']; 

$url = 'http://vimeo.com/api/rest/v2'; 
$url .= '?'; 
$url .= 'method=vimeo.videos.search&'; 
$url .= 'oauth_consumer_key='.$api_key2.'&'; 
$url .= 'per_page=10&'; 
$url .= 'query='.$video_id.'&'; 
$url .= 'sort=relevant&'; 
$url .= 'full_response=1'; 

$ch = curl_init(); 
curl_setopt($ch, CURLOPT_HEADER, 0); 
curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1); 
curl_setopt($ch, CURLOPT_URL, $url); 
$curl_response = curl_exec($ch); 
curl_close($ch); 


$xmlObject = simplexml_load_string($curl_response); 

$title= 'title'; 
$id = 'id'; 
$video_url = 'videosurl'; 
$thumbnails = 'thumbnail[1]'; 
$outputXML .= ""; 

$outputXML .= "<rsp>\n"; 

foreach($xmlObject->videos as $video) { 
    $videoTagBegin = "\t<video>\n"; 
    $urlXML = "\t\t<videosurl>".$video->attributes()->$video_url."</videosurl>\n"; 
    $titleXML = "\t\t<title>".$video->attributes()->$title."</title>\n"; 
    $thumbXML = "\t\t<thumbnail>".$video->attributes()->$thumbnails."</thumbnail>\n"; 
    $videoTagEnd = "\t</video>\n"; 
    $outputXML .= $videoTagBegin.$titleXML.$thumbXML.$urlXML.$videoTagEnd; 
} 
$outputXML .= "</rsp>"; 

print $outputXML; 

?> 

任何幫助，將不勝感激。我完全卡住了。

這是我的JS它是否有助於在所有....

function closeDivs(e) { 
    e.preventDefault(); 
    $('').empty(); 
}; 

$(document).ready(function() { 

    $('#searchbtn').bind('click' || 'enter',function(e) { 
     if ($.trim($('#searchBox').val()) !== '') { 
      $('#videos').empty(); 
      closeDivs(e); 
      $('#videos').append('<img src="img/loading.gif" alt="loading" class="loading" />'); 
      getEvents(e); 
      getVideos(e); 
     } 
    }); 

function getEvents(e) { 
    e.preventDefault(); 
    var text = 'event_id='+$('#searchBox').val(); 
    $.ajax({ 
     url: 'getEvents.php', 
     dataType: 'xml', 
     type: 'POST', 
     data: text, 
     success: function(data) {     
      }, 

     error: function(data) { 
       console.log('Error: ' + data); 
     } 

    }) 
}; 

function getVideos(e) { 
    e.preventDefault(); 
    var text = 'video_id='+$('#searchBox').val(); 
    $.ajax({ 
     url: 'getVideos.php', 
     dataType: 'xml', 
     type: 'POST', 
     data: text, 
     success: function(data) {        

     }, 
     error: function(data) { 
       console.log('Error: ' + data); 
     } 

    }) 
}; 
}); 

Answer 1

看來你得到什麼，從捲曲 回來嘗試amening你這樣的代碼，看看是否有任何錯誤

if($curl_response===false) 
{ 
    exit('Curl error: ' . curl_error($ch)); 
} 

之前關閉卷曲通話 - 看看到底是怎麼回事